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Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer Answers to Homework Set 9 From lecture 34: What is the rate constant at 298 K for a unimolecular reaction that has a free-energy of activation of 21 kcal mol–1? Now, rearrange eq. 34-4 so that all terms involving the variables t and [A] are on opposite sides of the equation. Integrate both sides from the starting point t = 0, [A] = [A]0 to t = t´, [A] = [A]t´ to obtain an expression for the concentration [A]t´ relative to the starting concentration [A]0 at any time t´. Using this equation, what is the half-life (the time required for [A]t´ to equal ½[A]0) for a unimolecular reaction as a function of the rate constant k? What is the half-life for the specific case of the unimolecular reaction having a free energy of activation of 21 kcal mol–1 (which is about right for rotation about the C–N bond in an amide, for instance)? We must evaluate k T k = B e"(GTS"GA )/RT h for ΔG = 21 in units of kcal mol−1. In that case, we must express R in units of kcal mol−1 K−1 and our exponential is ! e"(21)/(0.0019872#298) = 3.96 #10"16 and we have ! k T k = B " 3.96 "10#16 h −23 −1 −34 using kB = 1.3806 x 10 J K , T = 298 K, and h = 6.626 x 10 J s, we derive 2.46 x –3 −1 10 s . ! The requested rearrangement of eq. 34-4 gives d[A] " = kdt [A] If we integrate both sides over the indicated limits we have ! HW9-2 [A] d[A] t # " t # = k dt $ A $0 [ ]0 [A] A [ ]t # t # "ln([A]) = kt 0 [A]0 % [A] ( -+ % [A] ( -/ ln' 0 * = kt # ,note ln' 0 * = ln [A] " ln [A] 0 ' A * ' A * ( 0) ( t # ) & [ ]t # ) .- & [ ]t # ) 1- solving for the half-life time at which [A]t´ = ½[A]0 we have ! " % $ [A] ' ln 0 = kt ( $ 1 ' $ [A] ' # 2 0 & ln2 = kt ( ln2 = t ( k So, independent of starting concentration, the half life of a unimolecular reaction is the natural logarithm of 2 divided by the rate constant k. For the particular case of, say, an amide rotation having a ba!rr ier of 21 kcal mol−1, we can use the rate constant derived above to predict a half life of 282 seconds, or a bit less than 5 minutes. From lecture 35: Use the virial theorem (described in the solved homework of Lecture 16) to prove that since H2 is bound (i.e., is lower in energy than two separated H atoms) then the kinetic energy of the electrons in H2 must be greater than it is in two separated H atoms. For any system we have E = T +V where E is the total energy, T is the kinetic energy, and V is the potential energy. For electronic structure problems, the virial theorem says that ! V = –(2T) If we make this substitution above, we have ! E = –T ! HW9-3 So, if total energy goes down (becomes more negative), then kinetic energy must increase. Since bonding is a situation where energy goes down, the point is proven. Q.E.D. .

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